If A - y-y-log2-x-2- x-5- x is a whole number-B - y-y-x-2-x-0-1-y- C - 1- 3- 5- then n A- B - B - C is

Log_2|x 2| is not defined for x=2 now for x=0,1,3,4 |x 2|=2,1,1,2 ∴ A={0,1} y=|x 2|, x∈ [0,1] ⇒ y∈{1,2} (∵ y∈ℕ) ⇒ B={1,2} nbsp; C={1,3,5} (A× B) ∩ (B × C) ...

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Byju's Answer

Standard XII

Mathematics

Number of Elements in a Cartesian Product

If A = y:y=lo...

Question

If $A = {y : y = {log}_{2} | x - 2 |, x < 5, x is a whole number}, B = {y : y = | x - 2 |, x \in [0, 1], y \in N}, C = {1, 3, 5}$
, then $n ((A \times B) \cap (B \times C))$ is

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Solution

The correct option is C $1$
${log}_{2} | x - 2 |$ is not defined for $x = 2$
now for $x = 0, 1, 3, 4$
$| x - 2 | = 2, 1, 1, 2$
$∴ A = {0, 1}$
$y = | x - 2 |, x \in [0, 1] \Rightarrow y \in {1, 2} (∵ y \in N) \Rightarrow B = {1, 2} C = {1, 3, 5}$

$(A \times B) \cap (B \times C) = (A \cap B) \times (B \cap C)$
$= {(1, 1)}$

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Similar questions

Q. If

A = {y : y = {log}_{2} | x - 2 |, x < 5, x is a whole number}, B = {y : y = | x - 2 |, x \in [0, 1], y \in N}, C = {1, 3, 5}

, then

n ((A \times B) \cap (B \times C))

Q. Let

A = {y : y = {log}_{2} x, x < 16, x, y \in N}

and

B = {x : x^{2} - 7 x + 12 = 0} .

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(A \cup B) \times (A \cap B)

Q. If

A = {y : | y | = 3 x, x \leq 2, x \in N},

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and

C = {1, 4, 8},

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n ((A \times B) \cap (A \times C))

Q. If

A = {y : y = \frac{a + 1}{2}, a ϵ W and a \leq 5}, B {y : y = \frac{2 n - 1}{2}, n ϵ W and n < 5} and C = {- 1, \frac{- 1}{2}, 1, \frac{3}{2}, 2}

then

A - (B \cup C) =

Q. For

A = {x | - 3 \leq x < 4, x \in R}, B = {x | x < 5, x o n N}

and C= { 5, 3, 1, 0, 1, 3}, show that

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If A={y:y=log2|x−2|,x<5,x is a whole number},B={y:y=|x−2|,x∈[0,1],y∈N},C={1,3,5} , then n((A×B)∩(B×C)) is

The correct option is C 1log2|x−2| is not defined for x=2 now for x=0,1,3,4 |x−2|=2,1,1,2 ∴A={0,1} y=|x−2|, x∈[0,1]⇒y∈{1,2} (∵y∈N)⇒B={1,2}C={1,3,5} (A×B)∩(B×C)=(A∩B)×(B∩C) ={(1,1)}

If $A = {y : y = {log}_{2} | x - 2 |, x < 5, x is a whole number}, B = {y : y = | x - 2 |, x \in [0, 1], y \in N}, C = {1, 3, 5}$
, then $n ((A \times B) \cap (B \times C))$ is